int_{5 pi}^{25 pi}|sin 2 x+cos 2 x| d x= (in | vedclass

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Question

(C) Let $I = \int_{5 \pi}^{25 \pi} |\sin 2x + \cos 2x| dx$. We know that $\sin 2x + \cos 2x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2x + \frac{1}{\s

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) Let $I = \int_{5 \pi}^{25 \pi} |\sin 2x + \cos 2x| dx$. We know that $\sin 2x + \cos 2x = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin 2x + \frac{1}{\sqrt{2}} \cos 2x ight) = \sqrt{2} \sin(2x + \frac{\pi}{4})$. So,$I = \int_{5 \pi}^{25 \pi} |\sqrt{2} \sin(2x + \frac{\pi}{4})| dx = \sqrt{2} \int_{5 \pi}^{25 \pi} |\sin(2x + \frac{\pi}{4})| dx$. Let $2x + \frac{\pi}{4} = t$,then $2 dx = dt$,so $dx = \frac{dt}{2}$. When $x = 5\pi$,$t = 10\pi + \frac{\pi}{4} = \frac{41\pi}{4}$. When $x = 25\pi$,$t = 50\pi + \frac{\pi}{4} = \frac{201\pi}{4}$. $I = \frac{\sqrt{2}}{2} \int_{41\pi/4}^{201\pi/4} |\sin t| dt$. The period of $|\sin t|$ is $\pi$. The length of the interval is $\frac{201\pi}{4} - \frac{41\pi}{4} = \frac{160\pi}{4} = 40\pi$. Since the integral of $|\sin t|$ over one period $[0, \pi]$ is $\int_0^{\pi} \sin t dt = 2$,the integral over $40$ periods is $40 	imes 2 = 80$. Thus,$I = \frac{\sqrt{2}}{2} 	imes 80 = 40\sqrt{2}$.

$\int_{5 \pi}^{25 \pi}|\sin 2 x+\cos 2 x| d x=$ (in $\sqrt{2}$)

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